Question

Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.

Answer 1

\[ \text{ Given that } \]
\[ \text{ normal vector} , \vec{n} = \hat{i}  \]
\[\text{ Now } , \hat{n}  = \frac{\vec{n}}{\left| \vec{n} \right|} = \frac{\hat{k} }{\left| \hat{k} \right|} = \frac{\hat{k}}{1} = \hat{k} \]
\[\text{ The equation of a plane in normal form is } \]
\[ \vec{r} . \hat{n}  = d ( \text{ where d is the distance of the plane from the origin } )\]
\[ \text{ Substituting } \hat{n}  = \hat{k} \text{ and } d= 3 \text{ in the relation, we get } \]
\[ \vec{r} . \hat{k}  = 3\]